How much of a weak acid actually ionizes?
You will be able to: Set up a weak-acid equilibrium and check a small-change approximation.
How much of a weak acid actually ionizes?
Two bottles can both say 0.100 M acid yet have different pH values. A weak acid retains many undonated HA molecules; its hydronium concentration must be found from equilibrium.
A useful starting point: Why can one mole of a base supply two moles of hydroxide? →
Words and symbols before equations
- HA and A⁻
- A generic neutral monoprotic acid and its conjugate base.
- Ka
- Acid dissociation constant for HA + H₂O ⇌ H₃O⁺ + A⁻.
- ICE table
- Initial, change and equilibrium concentration bookkeeping.
- Percent ionization
- Equilibrium amount ionized divided by initial acid amount, times 100.
What this picture assumes
Dilute ideal-solution concentration model at 25 °C, Kw=1.00×10⁻¹⁴. Concentrations are mol/L (M); displayed values are rounded. No household experiments are required. A single monoprotic family with no added common ion. Charge balance includes water; percent ionization is the fraction converted to the conjugate form.
Read the picture in three steps
- Read the species and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- Weak acid: C=0.1 M, Ka=1.00e-5. pH=3.002, ionized fraction=1.00%. Both ionization and water are included.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
Starting with C M HA and negligible conjugate base, let x M ionize. The changes are −x for HA, +x for H₃O⁺ and +x for A⁻. Pure water is omitted from Ka.
Neglecting water’s small contribution gives Ka=x²/(C−x). If x is small relative to C, approximate C−x≈C and x≈√(KaC). A common classroom check is x/C≤5%; if it fails, solve the quadratic.
The positive quadratic root is x=[−Ka+√(Ka²+4KaC)]/2. The physical root must lie between zero and C. The explorer uses charge balance including water, so near-neutral cases remain physical.
Dilution usually raises the fraction ionized for a weak acid even while hydronium concentration falls. Fraction ionized and absolute concentration answer different questions.
A worked example, step by step
A weak acid has C=0.100 M and Ka=1.0×10⁻⁵. Estimate pH and percent ionization at 25 °C.
- Set Ka=x²/(0.100−x).
- Try x≈√(10⁻⁵×0.100)=1.0×10⁻³ M.
- Check x/C=0.010=1.0%, below the 5% guide.
- pH≈3.00 and ionization≈1.0%; the quadratic gives about 9.95×10⁻⁴ M, consistent with the estimate.
A small Ka alone does not justify an approximation; compare the calculated change with the initial concentration.
If x/C=20%, should C−x be replaced by C?
Compare with an explanation
No. A 20% change is too large for the usual small-change approximation.
Predict. Change one thing. Explain.
Choose weak acid and pK=5. Compare C=0.100 M with 0.010 M. Predict pH and percent ionization separately; explain their opposite trends.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Weak acid: C=0.1 M, Ka=1.00e-5. pH=3.002, ionized fraction=1.00%. Both ionization and water are included.
Dilute ideal-solution concentration model at 25 °C, Kw=1.00×10⁻¹⁴. Concentrations are mol/L (M); displayed values are rounded. No household experiments are required. A single monoprotic family with no added common ion. Charge balance includes water; percent ionization is the fraction converted to the conjugate form.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using proton transfer, charge and atom conservation, a mole balance or the stated acid–base equilibrium. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionFor C=0.010 M and Ka=1.0×10⁻⁴, test the square-root approximation and state a more suitable equation.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: The estimate is x=√10⁻⁶=0.0010 M.
- 1 point: x/C=10%, exceeding the usual 5% guide.
- 1 point: Use x²/(0.010−x)=1.0×10⁻⁴ instead of replacing the denominator.
- 1 point: The positive root is about 9.51×10⁻⁴ M, giving pH about 3.02.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What does x represent?
Concentration of HA that ionizes in the simple ICE setup.
RECALL 2Why omit water from Ka?
Its pure-liquid activity is treated as constant.
RECALL 3What must be checked after approximation?
The fractional change x/C.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How much of a weak acid actually ionizes?
- Ka≈[H₃O⁺][A⁻]/[HA].
- If valid: x≈√(KaC); percent ionization=100x/C.
Remember: A small Ka alone does not justify an approximation; compare the calculated change with the initial concentration.
Conditions: Dilute ideal-solution concentration model at 25 °C, Kw=1.00×10⁻¹⁴. Concentrations are mol/L (M); displayed values are rounded. No household experiments are required. A single monoprotic family with no added common ion. Charge balance includes water; percent ionization is the fraction converted to the conjugate form.
Refresh Kid · AP Chemistry Unit 8 · Objectives 8.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 8.3, objective 8.3.A. CED effective Fall 2024 and June 2026 clarifications checked September 17, 2026. Unit 8: Acids and Bases, Topics 8.1–8.11. Focused lesson names, examples, models and assessments are original Refresh Kid teaching materials, not additional official topics or official AP questions. Official corrections.
The model states its assumptions beside the diagram. Dilute ideal-solution concentrations approximate activities; numerical models use 25 °C and Kw=1.00×10⁻¹⁴ unless another pKw is supplied. pH need not be restricted to 0–14 in all real solutions. The optional 3D views show original schematic molecular geometry, not a measured trajectory or a reaction mechanism. Computation of a buffer’s pH change after adding acid/base, derivation of Henderson–Hasselbalch, concentrations of every species in a polyprotic titration, and solubility as a function of pH are excluded from assessed scope. Buffer response and pH-dependent solubility are taught qualitatively. Calculating the pH of a buffer formed by partial neutralization remains in Topic 8.4 scope.
Teaching resources: The Organic Chemistry Tutor video titles/descriptions and topic coverage were checked for optional links; no claim is made to have watched every video. No creator scripts, examples, worksheets or artwork were copied. GitHub’s 3D website collection and its Three.js camera-control example informed the idea of controllable spatial inspection. Scientific diagrams, geometry and interactions here are original. The self-hosted Three.js runtime retains its MIT license. Camera rotation changes the view, not the chemistry.
Independent teacher review and observation of students remain pending. Implementation checks do not certify scientific accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Chemistry questions and scoring guides. This archive contains questions across units; it is not an assignment of every question to this lesson.
The teaching sequence is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
Want to work through this with a tutor?
Bring your question about How much of a weak acid actually ionizes? Your explanation and answers remain free to access.